SOLUTION: evaluate log3 square root of 27 over 9 + log2 16 square root of 32

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Question 231097: evaluate log3 square root of 27 over 9 + log2 16 square root of 32
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%283%2C+%28sqrt%2827%29%2F9%29%29+%2B+log%282%2C+%2816%2Asqrt%2832%29%29%29
The easy way to solve this is to recognize that 27 and 9 are powers of 3 and that 16 and 32 are powers of two. This will make finding the two logs easy. If we don't recognize this, then we are looking at changing the base of the logarithms to a base you have on your calculator and then using the calculator to find the answer.

But we're going to do the problem the easy way.
27+=+3%5E3
9+=+3%5E2
16+=+2%5E4
32+=+2%5E5
and exponents of 1/2 mean square root. Using all these our expression becomes:

Using our rules for exponents this simplifies:

log%283%2C+%283%5E%28%28-1%29%2F2%29%29%29+%2B+log%282%2C+%282%5E%2813%2F2%29%29%29
Now we can use a property of logarithms, log%28a%2C+%28x%5Ey%29%29+=+y%2Alog%28a%2C+%28x%29%29, to "move" the exponents out of the arguments:
%28-1%2F2%29log%283%2C+%283%29%29+%2B+%2813%2F2%29log%282%2C+%282%29%29
Now, by definition:
log%283%2C+%283%29%29++=+1 and
log%282%2C+%282%29%29+=+1 so our expression becomes:
%28-1%2F2%29%2A1+%2B+%2813%2F2%29%2A1
which simplifies...
%28-1%2F2%29+%2B+%2813%2F2%29
12%2F2
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